Question

# Consider the following data: ΔfH0 (N2H4,l) = 50 kJ/mol ΔfH0 (NH3,g) = −46k J/mol B.E.(N−H) = 393 kJ/mol and B.E.(H−H) = 436 kJ/mol, also ΔvapH(N2H4,l) = 18 kJ/mol The N- N bond energy in N2H4 is

A

226 kJ/mol

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B

154 kJ/mol

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C

190 kJ/mol

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D

45.45 kJ/mol

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Solution

## The correct option is C 190 kJ/mol 12N2 + 32H2 → NH3 (g) (g) (g) Let BF of N≡ N s x −46 = x2 + 32 × 436 − 3 × 393 ∴ x = 958 N2H4(I) → N2(g) + 2H2(g) ΔH = −50kJ/mol ΔH = [ΔHvap(N2H4,l) + 4 × BE(N − H) + BE(N − N) + BE(N − N)] − [B.E of (N ≡N)] + 2BE(H − H) − 50 = (18 + 4 × 393 + y) − [958 + 2 × 436] 50 = (1590 + y) − (1830) BE of (N − N) (or) Y = 190 KJ/mol

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